Certainly! Let's consider the statement made by Lindsay and analyze it step-by-step.
1. Identifying two consecutive odd integers
- Let's denote the first odd integer by [tex]\( z \)[/tex].
- The next consecutive odd integer would then be [tex]\( z + 2 \)[/tex].
2. Calculating the product of the two odd integers
- The product of these two integers can be expressed as [tex]\( z \times (z + 2) \)[/tex].
- Simplifying this expression: [tex]\( z \times (z+2) = z^2 + 2z \)[/tex].
3. The even integer between these two odd integers
- The even integer between [tex]\( z \)[/tex] and [tex]\( z + 2 \)[/tex] is [tex]\( z + 1 \)[/tex].
4. Calculating the square of the even integer minus 1
- The square of the even integer is [tex]\( (z + 1)^2 \)[/tex].
- Expanding this expression: [tex]\( (z + 1)^2 = z^2 + 2z + 1 \)[/tex].
- Now, subtract 1 from this squared value: [tex]\( (z^2 + 2z + 1) - 1 = z^2 + 2z \)[/tex].
5. Comparing both expressions
- The product of the two odd integers, as shown above, is [tex]\( z^2 + 2z \)[/tex].
- The square of the even integer minus 1 simplifies to [tex]\( z^2 + 2z \)[/tex].
- Therefore, we see that these two expressions are indeed equal.
Thus,
Two consecutive odd integers can be represented as [tex]\( z \)[/tex] and [tex]\( z + 2 \)[/tex]. The product of the two integers is [tex]\( z^2 + 2z \)[/tex].
The even integer between the odd integers is represented by the expression [tex]\( z + 1 \)[/tex].
The squared expression minus 1 is [tex]\( z^2 + 2z \)[/tex].
Therefore, the product of the consecutive odd numbers is indeed equal to the square of the even integer minus 1.
Lindsay's statement is true.
